Basic trigonometry in dynamics force problem

[Bunny-chan]

Homework Statement

Two spheres, with $0.5g$ each, are hanging by $30cm$ threads, tied on the same spot. The same electric charge is communicated to each sphere; in consequence, the threads move apart until they are about $60^\circ$ from each other. What is the value of the charge?

$\theta = \frac{60^\circ}{2} = 30^\circ \\ \\m = 0.5g = 0.0005kg\\ d = 30cm = 0.3m$

Homework Equations

Not needed.

The Attempt at a Solution

So, what I did was the following:

$$\tan \theta = \frac{\vec F}{\vec P} \\ \Rightarrow \frac{\sqrt 3}{3} = \frac{\vec F}{m\vec g} = \frac{\vec F}{0.0005 \times 9.8} = \frac{\vec F}{0.005} \\ \Rightarrow \vec F = \frac{\sqrt 3}{3} \times 0.005 = 0.0028 \\$$$$k \frac{q^2}{d^2} = 0.0028 \\ \Rightarrow 9 \times 10^9 \times \frac{q^2}{0.3^2} = 2.8 \times 10^{-3} \\ \Rightarrow q^2 = \frac{(2.8 \times 10^{-3}) \times (9 \times 10^{-2})}{9 \times 10^9} = 2.8 \times 10^{-14} \\ \Rightarrow q = 1.7 \times 10^{-7}C$$My result did match the answer in the textbook, but searching through the web, I came across a different way of solving it:

The rest is just like mine, so I didn't bother putting it. Anyway, he reached the same result, but with a different method and trigonometric relations, which I'm quite lacking in and I couldn't understand very well yet.

Why did he conclude that

and

?

I know I could just go on and ignore this because I already solved the exercise, but vectors are a puzzling topic for me and I feel like there's something I'm missing... I'd appreciate some help!

 

[patric44]

i hope that help

 

[patric44]

the d part :

 

[Bunny-chan]

i hope that help
View attachment 197535

the d part :
View attachment 197536

That's perfect, thank you!